Statics of Building Structures I., ERASMUSfast10.vsb.cz/koubova/SoBSI_theme4_three_mom_eq.pdfStatics...

Department of Structural Mechanics

Faculty of Civil Engineering, VŠB-Technical University of Ostrava

Statics of Building Structures I., ERASMUS

Continuous beam

• Basic properties of a continuous beam

• Solution of a continuous beam by the Force method

• Symmetry of a continuous beam

2 / 40Basic properties of a continuous beam

Supports of the transversally loaded

continuous beam

Continuous beam

2) 1, (0, supports fixed ofnumber

spans ofnumber

n

:acyindetermin statical of Degree

support) (fixed

node endat rotation against b)

movement, rticalagainst ve a)

: restrained isIt

loading. sversal with tranbeamdirect ateindetermin statically a is beam Continuos

s

k

k

v

p

vp 1

3 / 40Solution of a continuous beam by the Force method

First 3 steps of the Force method in

solution of a continuous beam

Continuous beam, derivation of the “Three moments equation”

.conditions naldeformatio writing4)

supports), fixed(at reactionsor nsinteractio

moment with links removed of replacing 3)

hinges), (inserting links internal of removal 2)

acy indetermin statical of degree ofion determinat 1)

:equation moments Three derive tosteps basic method, Force The

s

s

n

n ,


Derivation of the Three moments equation

1,1, :condition nalDeformatio rrrr

5 / 40

Calulation of rotations at the ends of beams

Simply supported beam as a member of a statically indeterminate structure

Deformations of a simply supported beams at the ends

abababbaababbaabbabababa MMMM ,,,,0,,,,,,,0,,,

:rotation clockwise For the

6 / 40

Derivation of 3moments (Clapeyron’s) equation

equation each in moments bendingunknown 3maximally are there

,acy indetermin statical of degree its toequal is beam continuous afor equations ofnumber

rotation), (clockwise #5 and #4 slideson picturesat marking toscorrespond ofsign

:Notes

:conversion andon substitutiAfter

,1rr,ba,

,1rr,ba,

,1rr,

:is 1)r (r,span right For

,1-rr,ab,

,1-rr,ab,

,1-rr,

:is r) 1,-(rspan left For

sn

rr

rrrrrrrM

rrrrrM

rrrM

rrrrM

rrrM

rrrr

rM

abM

rM

baM

ba

rrrM

rrrrM

rrrr

rM

baM

rM

abM

ab

rrrr

0,1,

00,1,0,1,1,1

)1,1,

(1,1

1,1,1,0,1,1,

1,,

,,

1,1,,10,1,1,

1,,

,,

1,1,


Derivation of the Three moments equation

:support For

:is beam theof endright For

:is beam theof endleft For

.( is equations ofnumber

ate,indetermin statically times1)-(p is (b) and (a) pictures ofconnetion by obtained beam continuousA

:composed be illequation w 1Only

1.ancy indetermin statical of degree has (c) below picture at the beam The

0)(

0)(

0)(

)1,0

0)(

.,0,0

0,1,0,1,1,11,1,,11

0,1,0,1,1,1,,11

0,1,20,3,23,233,21,22

11

0,1,20,3,21,22,12

231

rrrrrrrrrrrrrrr

pppppppppppp

p

rs

MMMr

MM

MM

pMM

M

MMMMn


Derivation of the 3moments equation, the beam with a cantilever

s.cantilever theof loading fromdirectly calculated becan moments Those

below) depicted case in the negative are(they

loaded are ends cantilever when zero-non are supportsouter above moments Bending


Derivation of the 3moments equation, fixed end

0,0

0

"

0

0,2,12,10,,1,11,1

0,0,10,1

0,2,12,122,11

,12,1

pppppppppppp

pp

MM

MM

2p1,p

1,0

:(d)) (c), (picture end fixedright for Similarily

is Here

picture).in pole" nulové (ie. span" zero" called-so inserted with (b) picturefor validiscondition same The

:is (a)) (picture end fixedfor condition nalDeformatio

equation. moments Three theof solvingby calculated becan moments The

zero.-non moments bendingbut zero are , ends fixedat Rotations

10 / 40

Variable cross-section within span

0)(2

6,

6

0)(6)(2

6,

633

0)(

1,

1,

0,1,

1,

1,

0,1,

1,

1,

1

,1

,1

1,

1,

,1

,1

1

0,1,

1,

1,

0,1,

1,

1,

0,1,0,1,

1,

1,

1

,1

,1

1,

1,

,1

,1

1

1,

1,

1,

1,

11,

1,

1,

1,

0,1,0,1,1,1,11,,11

rr

rr

rr

rr

rr

rr

rr

rr

r

rr

rr

rr

rr

r

rr

rr

r

rr

rr

rr

rr

rr

rr

rrrr

rr

rr

r

rr

rr

rr

rr

r

rr

rr

r

rr

rr

rr

rr

rr

rr

rr

rr

rrrrrrrrrrrrrrr

l

lZ

l

lZ

I

lM

I

l

I

lM

I

lM

l

IE

l

IE

EI

lM

I

l

I

lM

I

lM

IE

l

IE

l

IE

l

IE

l

MMM

: then Z ZmarkingWhen

:conversionAfter

, ,

:spanevery in section -cross invariablefor is

equation In

1rr,1-rr,

1-rr,1rr,1-rr,1rr,

11 / 40

3moments equation – constant cross-section on all beam

Assumption:

materially and geometrically invariable cross-section

on all beam, ie. E·I = konst.

Then the Three moments equation has form:

0)(2 1,1,1,1,1,11,1,1,1 rrrrrrrrrrrrrrrrrrr lZlZlMllMlM


Loading members

Formulas for loading members of

the Three moments equation

beam supportedsimply aon determined is

a Rotation

:loading thermaland forcefor members Loading

a,0b,b,0a,

0,,

,

,

,

0,,

,

,

,

6

6

ab

ba

ba

ab

ba

ba

ba

ba

l

IEZ

l

IEZ

13 / 40

Internal forces of the continuous beam

ba

bbaa

xbaaxx

ba

abxabxx

ba

ababababab

ba

abbabababa

ba

l

xMxlMMxVMMM

l

MMVVVV

l

MMVVVV

l

MMVVVV

MM

,

,

0,,0,

,

0,,0,

,

0,,,0,,,

,

0,,,0,,,

)(

: writtenbecan moments bendingFor

:forcesshear for can write We

. a momentsby b) beam, supportedsimply a as a)

:loaded isIt beam. continuous theofpart a is b-a beam The

14 / 40

Reactions of a continuous beam

1,1,

1

rrV

rrVrR

rR : writtenbecan r support aat reaction verticalaFor

sideright at 1rr,span theb)

sideleft at r ,-rspan thea)

:separatesr support The

15 / 40

Loading members for loading by displacement of supports

1p

1p

1

1

1,0rr,1,0rr,

: is end fixed at the 1)("support right theof rotationa clockwiseFor

: is end fixed at the "support left theofrotation clockwiseFor

:is )( nt displaceme alFor vertic

:span) 1ithin constant w(but spansin section -crossdifferent with beam continuousfor equation moments Three The

El

wwE

I

lM

I

lM

p

El

wwE

I

lM

I

lM

l

ww

l

wwE

I

lM

I

l

I

lM

I

lM

l

ww

l

ww

w

EI

lM

I

l

I

lM

I

lM

pp

pp

pp

pp

p

pp

pp

p

rr

rr

rr

rr

rr

rr

r

rr

rr

rr

rr

r

rr

rr

r

rr

rr

rr

rr

r

rrrr

rr

rr

r

rr

rr

rr

rr

r

rr

rr

r

6)(62

662

0)(6)(2

0)(6)(2

1.

1

1,

1,

1

1,

1,

1,2

12

2,1

2,1

2

2,1

2,1

1

1,

1

1,

1

1,

1,

1,

1,

,1

,1

,1

,1

1

1,

1

1,

1

0,1,0,1,

1,

1,

1,

1,

,1

,1

,1

,1

1

"

"1


Problem definition and solution

of the example 4.1 (part one)

Example 4.1

loading Thermal

(a), picture toaccording loading Force

:definition Problem

mhmhh

mImII

6,03,0

1040,106

3,24,32,1

44

3,2

44

4,32,1


Solution of example 4.1 (part two)

Example 4.1, displacement of supports

kPaEm,m , l,,hmI

m,lm,,m, l,h,hmII

)mm(,w), mm(w

,,

-

,

,,,,

-

,,

7

3232

44

32

43214321

44

4321

32

104,246601040

138230106

532

,

:definition Problem

18 / 40

Example 4.1, displacement of supports, solution

.10129032,1101,3

05,3,10234375,010

4,6

25,3

,10234375,0104,6

25,3,10714286,010

6,2

02

,10714286,0108,2

02

,,,,,

0)(6)(2

0)(6)(2

062

33

0,4,3

33

0,2,3

33

0,3,2

33

0,1,2

33

0,2,1

4,3

340,4,3

3,2

230,2,3

3,2

230,3,2

2,1

120,1,2

2,1

120,2,1

0,4,30,4,3

4,3

4,3

3,2

3,2

3

3,2

3,2

2

0,1,20,3,2

3,2

3,2

3

3,2

3,2

2,1

2,1

2

2,1

2,1

1

0,2,1

2,1

2,1

2

2,1

2,1

1

:onsubstitutiafter

:isIt

:equations ofn Compositio

l

ww

l

ww

l

ww

l

ww

l

ww

EI

l

I

lM

I

lM

EI

lM

I

l

I

lM

I

lM

EI

lM

I

lM

19 / 40

Example 4.1, displacement of supports, solution continuation

kNVRkNVVR

kNVVRkNVR

kNl

MMVVkN

l

MMVV

kNl

MMVV

kNmMkNmMkNmM

MMM

MMM

MMM

MM

MMM

MM

309,4,877,4309,4568,0

,576,8568,0144,9144,9

309,41,3

0358,13568,0

4,6

722,9358,13

144,98,2

722,9881,15

,358,13,722,9,881,15

6,1958983,1353316000

2,6910716003,125337,4666

041,2202

010)234375,0129032,1(104,26)106

1,3

1040

4,6(2

1040

4,6

010)714286,0234375,0(104,261040

4,6)

1040

4,6

106

8,2(2

106

8,2

010714286,0104,26106

8,2

106

8,22

3,444,32,33

3,21,222,11

4,3

433,44,3

3,2

322,33,2

2,1

211,212

321

321

321

321

37

44342

37

4344241

37

4241

:

:

:úpravě Po

:

reactions and forcesShear

is equations 3 of system theofSolution

equations of system intoon Substituti



of the example 4.2

Example 4.2

21 / 40Employment of the symmetry of a continuous beam

Symmetry of geometry and supports of continuous beam

Symmetry of a continuous beam

)cantileveror fixed ended,roller or ended(pin

same are beam continuous a of endsboth - supports ofsymmetry b)

sections-cross identical andlenght identical have spans lsymmetrica -geometry ofsymmetry a)

:assumes beam continuous a ofSymmetry

spans ofnumber even b)

spans ofnumber odd a)

: withbeams lsymmetricafor symmetry of axis ofposition different is There


Symmetrical, antisymmetrical and

general loading

Loading of a symmetrical continuous beam

spans ofnumber even loading, ricalantisymmet AS

spans ofnumber even loading, lsymmetrica SS

spans ofnumber even loading, ricalantisymmet AL

spans ofnumber odd loading, lsymmetrica SL

:

ryantisymmetor symmetry

ofcharacter without loading c) ad

forces ofdirection oposite of pictures mirrored b) ad

forces ofdirection same of pictures mirrored a) ad

general c)

A- ricalantisymmet b)

S - lsymmetrica a)

:

onsAbbraviati

:by formed

is sidesboth on Loading

becan beam continuous lsymmetrica a of loading The


Utilization of a symmetry for symmetrical and antisymmetrical

loading of a continuous beam.

Symmetrical and antisymmetrical loading

2

1 , ´,:AS ,

2 ́ , ́ :AL

2

1 , ´, :SS ,

2 ́ , ́ :SL

3223322

3223322

sAS

s

sAL

s

sSS

s

sSL

s

nnMMM

nnMMMM

nnMMM

nnMMMM



of the example 4.3 (part one)

Example 4.3

beam. lsymmetrica

same theof states loading ricalantisymmet

and lsymmetrica theof solutions of

ionsuperpositby given issolution Final

moments..) bending (shear, forces internal of

evaluation including ,separately solved are

beam continuous a of states loadingBoth

loading ricalantisymmet b)

loading symetrical a)

:into decomposed becan beam

continuous lsymmetrica a of Loading


Solution of example 4.3 (part two)

Example 4.3

Resulting diagrams of internal forces - by superposition of SL+AL

26 / 40

Donau-wald bridge, Winzer, Germany

Examples of real continuous beam structures

27 / 40

Donau-wald bridge, Winzer, Germany


28 / 40

Bogenberg bridge, Bogen, Germany


29 / 40

Kingstone Bridge, Glasgow, Scotland


30 / 40



31 / 40



32 / 40

Bridge in Nusle quarter, Prague


33 / 40

Bridge in Nusle quarter, Prague


34 / 40

Construction of highway D47, Ostrava


35 / 40



36 / 40



37 / 40

Energetic Research Centre, VŠB-TU Ostrava


38 / 40



39 / 40



40 / 40



Statics of Building Structures I., ERASMUSfast10.vsb.cz/koubova/SoBSI_theme4_three_mom_eq.pdfStatics...

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